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3 years ago

Find the value of 9-6(8-5)+(-3)

Mathematics

1 answer:

mestny [16]3 years ago

5 0

Step 1: Distribute

9-48+30-3

Step 2: Solve

9-48+30-3=-12

Your answer is -12.

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What is the equation of the line that passes through the point (-4, -6) and has a slope of -1/2?

Sedaia [141]

Point slope form

y - -6 = (-1/2)(x - -4)

y + 6 = (-1/2)x - 2

y = (-1/2)x - 8

Answer: y = (-1/2)x - 8

Check. Clearly slope is -1/2, good. (-1/2)(-4)-8=2-8=-6 good

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4 years ago

Paul's car is 17 feet long. He is making a model of his car that is the actual size. What is the length of the model?

amid [387]

Answer:

Step-by-step explanation:

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3 years ago

Type the correct answer in each box. Use numerals instead of words.

Alex_Xolod [135]

Answer:

Step-by-step explanation:

Given that prices for a pair of shoes lie in the interval

[80,180] dollars.

Delivery fee 20% of price.

i.e. delivery fee will be in the interval [4, 9]

(1/20th of price)

Total cost= price of shoedelivery cost

Hence f(c) = c+c/20 = 21c/20

The domain of this function would be c lying between 80 to 180

So domain =[80,180]

---------------------------------

Amount to be repaid = 42 dollars

Once he received this amount, the price would be

105+42 =147

But since price range is only [21*80/20, 21*180/20]

=[84, 189]

Since now Albert has 147 dollars, he can afford is

[80,147]

6 0

4 years ago

Read 2 more answers

Which is a solution to (x-3)(x 9)=-27

dezoksy [38]

Distribute
(x-3)(x+9)=-27
x^2+6x-27=-27
add 27 to both sides
x^2+6x=0
factor out x
x(x+6)=0
set each to zero
x=0
x+6=0
x=-6

x=0 and or -6

x=-6 or 0

8 0

3 years ago

Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$

$S=10x^{2}-18x-15x+27$ (expand)

$S=10x^{2}-33x+27$ (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12+27$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x+15$

$20x^{2}-66x+27=0$

Solving:

$\frac{66+\sqrt{(-66)^{2}-(4.20.27)} }{2(20)}$

$\frac{66+46.86}{40}$

$x_{1}=\frac{66+46.86}{40}=$ 2.82

$x_{2}=\frac{66-46.86}{40}$ = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82

6 0

3 years ago